Question: $f(x, y, z) = xy - y^3 + \tan(z)$ What is $\text{div}(\text{grad}(f))$ ? $\text{div}(\text{grad}(f)) = $
Solution: The Laplacian of a scalar field $f$ is the sum of each of its second partial derivatives. $\text{div}(\text{grad}(f)) = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2}$ [What does it mean to take the divergence of the gradient?] Let's find the second partial derivatives of $f$ ! $\begin{aligned} f_{xx} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] = \dfrac{\partial}{\partial x} \left[ y \right] = 0 \\ \\ f_{yy} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] = \dfrac{\partial}{\partial y} \left[ x - 3y^2 \right] = -6y \\ \\ f_{zz} &= \dfrac{\partial}{\partial z} \left[ \dfrac{\partial f}{\partial z} \right] \\ \\ &= \dfrac{\partial}{\partial z} \left[ \dfrac{1}{\cos^2(z)} \right] \\ \\ &= \dfrac{0 - 2\cos(z)(-\sin(z))}{\cos^4(z)} \\ \\ &= \dfrac{2\sin(z)}{\cos^3(z)} \end{aligned}$ The Laplacian is $\text{div}(\text{grad}(f)) = f_{xx} + f_{yy} + f_{zz}$. Therefore: $\text{div}(\text{grad}(f)) = -6y + \dfrac{2\sin(z)}{\cos^3(z)}$